I should mention that I still haven't taken Calculus or even Pre-Calculus, which is why I want to ask this. I've seen proofs $e$ is irrational, but not this one. Is this correct, and if it isn't, why not?
Prove $e$ is irrational:
One definition for $e$ is $$ e = \lim\limits_{n\to \infty} (1+\frac{1}{n})^n $$
Let's assume $e$ is rational. Because of that, we can set: $$ (1+\frac{1}{n})^n = \frac{a_n}{b_n}$$ $n$-root both sides: $$ 1+\frac{1}{n} = \sqrt[n]{\frac{a_n}{b_n}} $$ Distribute the $n$-root: $$ 1+\frac{1}{n} = \frac{\sqrt[n]{a_n}}{\sqrt[n]{b_n}} $$ Multiply by $\sqrt[n]{b_n}$ on both sides: $$ \sqrt[n]{b_n}(1+\frac{1}{n}) = \sqrt[n]{a_n} $$ Distribute the $\sqrt[n]{b_n}$: $$ \sqrt[n]{b_n}+\frac{\sqrt[n]{b_n}}{n} = \sqrt[n]{a_n} $$ Add the fractions: $$ \frac{n\sqrt[n]{b_n}}{n}+\frac{\sqrt[n]{b_n}}{n} = \sqrt[n]{a_n} $$ $$ \frac{n\sqrt[n]{b_n}+\sqrt[n]{b_n}}{n} = \sqrt[n]{a_n} $$ Factor out $\sqrt[n]{b_n}$: $$ \frac{\sqrt[n]{b_n}(n+1)}{n} = \sqrt[n]{a_n} $$ Divide by $\sqrt[n]{b_n}$ on both sides: $$ \frac{(n+1)}{n} = \frac{\sqrt[n]{a_n}}{\sqrt[n]{b_n}} $$ Factor the $n$-root: $$ \frac{(n+1)}{n} = \sqrt[n]{\frac{a_n}{b_n}} $$ Raise both sides to the power $n$: $$ (\frac{(n+1)}{n})^n = (\sqrt[n]{\frac{a_n}{b_n}})^n $$ Distribute the power $n$: $$ \frac{(n+1)^n}{n^n} = \frac{a_n}{b_n} $$
So, by the transitive property of equality, we have:$$ \frac{a_n}{b_n} = (1+\frac{1}{n})^n = \frac{(n+1)^n}{n^n} $$
Substituting that in, we have: $$ e = \lim\limits_{n\to \infty} \frac{(n+1)^n}{n^n}$$ We can just put infinity into the equation: $$ e = \frac{(\infty + 1)^\infty}{\infty^\infty} $$ $$ e = \frac{\infty^\infty}{\infty^\infty} $$ $$ e = \frac{\infty}{\infty} $$ Which is indeterminate. Because of this, out original assumption that $e$ is rational was wrong, therefore, $e$ is irrational.
Your proof is not correct.
First note that (toward the end) $$ \lim_{n\to \infty} \frac{(n+1)^n}{n^n} = \lim_{n\to \infty} \left(1 + \frac{1}{n}\right)^n $$ So you haven't really done anything.
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Also, you say that since $e$ is rational $$ \left(1 + \frac{1}{n}\right)^n $$ is rational ($n\in \mathbb{Z}\setminus \{0\}$). But that is true even is $e$ is not rational.
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Just cause $\lim_{x\to \infty} f(x) = \infty$ and $\lim_{x\to \infty} g(x) = \infty$, you can't say anything about $\lim_{x\to \infty} \frac{f(x)}{g(x)}$.
For example:
$$ \lim_{n\to \infty} \frac{n}{n} = 1 $$ and $$ \lim_{n\to \infty}\frac{n^2}{n} = \infty $$